The points $(2, 1)$ and $(-3, -4)$ are opposite vertices of a parallelogram. If the other two vertices lie on the line $x + 9y + c = 0$, then $c$ is

We have a parallelogram where two opposite vertices are at $(2, 1)$ and $(-3, -4)$, and the other two vertices lie on the line $x + 9y + c = 0$. We need to find the value of $c$.

In any parallelogram, the diagonals bisect each other. Since $(2, 1)$ and $(-3, -4)$ are opposite vertices, they form one diagonal of the parallelogram.

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The midpoint of two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$

For points $(2, 1)$ and $(-3, -4)$:

$\text{Midpoint} = \left(\dfrac{2 + (-3)}{2}, \dfrac{1 + (-4)}{2}\right)

= \left(\dfrac{-1}{2}, \dfrac{-3}{2}\right)$

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Let's call the other two vertices $A$ and $B$. Since they're also opposite vertices, they form the second diagonal of the parallelogram.

Both diagonals intersect at the same point (the center of the parallelogram).

Since both vertices $A$ and $B$ lie on the line $x + 9y + c = 0$, and their midpoint is the center of the parallelogram, this center must also lie on the line.

Therefore, the point $\left(-\tfrac{1}{2}, -\tfrac{3}{2}\right)$ lies on the line $x + 9y + c = 0$.

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Since $\left(-\tfrac{1}{2}, -\tfrac{3}{2}\right)$ lies on the line $x + 9y + c = 0$:

$-\dfrac{1}{2} + 9\left(-\dfrac{3}{2}\right) + c = 0$

$-\dfrac{1}{2} - \dfrac{27}{2} + c = 0$

$-\dfrac{1 + 27}{2} + c = 0$

$-\dfrac{28}{2} + c = 0$

$-14 + c = 0$

Therefore: $c = 14$

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Answer: $c = 14$

In parallelogram problems, the center (intersection of diagonals) is always the midpoint of both diagonals. This property is the key to solving such problems efficiently.